# Sequence of Functions on $[0,1]$ with Derivatives Bounded by $L^1$ Function

I'm stuck on the last step of a real analysis/advanced calculus problem and could really use some help. The problem is as follows:

Let $f_n$ be continuously differentiable on $[0,1]$ satisfying, for all $n \geq 1$ ,

$$\hspace{.5cm} |f_n'(x)| \leq \frac{1}{\sqrt{x}}, \hspace{.3cm} 0 < x \leq 1, \hspace{.3cm} \text{and}$$ $$\int_0^1 f_n dx = 0.$$

Show that there exists a subsequence converging uniformly on $[0,1]$.

So naturally I want to use the Arzela-Ascoli Theorem, i.e. we need to show that the family is equicontinuous and uniformly bounded. By a standard fact, on a closed interval it's enough to show boundedness at a single point: Arzela-Ascoli Theorem: pointwise boundedness at one point

It's easy to do so at $x = 1$; by the derivative bound, comparing the function to $2\sqrt{x}$ we see that $f_n(1) \leq 2$ lest it be positive everywhere and thus have a non-zero integral. Similarly, if $f_n(1) < -2$ the function will be negative-everywhere. Thus the family is bounded at $1$, so it remains to show equicontinuity.

A similar argument can give equicontinuity on all intervals $[a,1]$ for $0 < a \leq 1$, but I have trouble completing the problem. So assuming this proven, the contradiction would be if there are $\delta_k \rightarrow 0$ and $f_k$ such that $|f_k(\delta_k) - f_k(0)| \geq \epsilon$ for some $\epsilon > 0$.

Can somebody help me with the last step? There are probably several equivalent proofs; I'd enjoy seeing the various ways to reason out the last bit!

Thanks!

Hint: As $f_n$ is continuously differentiable, you have $$f_n(x)=\int_0^x f_n^{\prime}(t)dt+f_n(0)$$ and hence, for $x,y\in [0,1]$ $$f_n(x)-f_n(y)=\int_y^x f_n^{\prime}(t)dt$$
• Haha yeah, that works. I don't know what my hold-up was; can also just do the comparison with $2\sqrt{x}$ at $0$ as well... Just brainfart. – John Samples Jul 5 '16 at 9:50